Subject: [mg125134] Re: Why the FourierTransform gives two different answers?

From: "Nasser M. Abbasi" <nma at 12000.org>

Date: Thu, 23 Feb 2012 05:48:07 -0500 (EST)

Delivered-to: l-mathgroup@mail-archive0.wolfram.com

References: <jhvupg$893$1@smc.vnet.net>

Reply-to: nma at 12000.org

On 2/21/2012 5:22 AM, Ð?Ð»ÐµÐºÑ?ÐµÐ¹ wrote:
> Why the FourierTransform gives two different answers?
>
> In[1] FourierTransform[ (t - 5.0)^2*Exp[-(t - 5.0)^2 ], t, w]
>
> In[2] FourierTransform[ (t - 5)^2* Exp[-(t - 5)^2 ], t, w]
>
It seems to be in integration part. In this below I used
the text book definition of the F.T. itself, and instead
of integrating to infinity, integrate to 'k', then take the limit
as k->Infinity of the result of the integration, and I get the
same result of both the symbolic and the numerical version.
So, the problem is with the limit to infinity with the numerical
version inside the integration code.
-----------------numerical ------------------
ClearAll[t, w, k]
f = (t - 5.0)^2*Exp[-(t - 5.0)^2];
res = Integrate[f*Exp[-I w t], {t, -k, k}];
res = Limit[res, k -> Infinity];
Plot[{Re[res], Im[res]}, {w, -5, 5}]
-------------------------
------------ symbolic -----------
ClearAll[t, w, k]
f = (t - 5)^2*Exp[-(t - 5)^2];
res = Integrate[f*Exp[-I w t], {t, -k, k}];
res = Limit[res, k -> Infinity];
Plot[{Re[res], Im[res]}, {w, -5, 5}]
-----------------------------
same result, as what one would expect.
If I change the numerical integration above to be
Integrate[f*Exp[-I w t], {t, -Infinity, Infinity}]
Then it become zero again. So, I think it is a bug
in the integration with the limit to infinity for this
integrand.
--Nasser